The image is the problem: ty :P

Fine the value of a in the image below.

telo118 [61]

Answer:

metre it with scale and then write its value with points

7 0

2 years ago

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My cats thoughts while she is just laying there with her face in my clothes:

nata0808 [166]

Answer:

ADORABLE!!

Step-by-step explanation:

7 0

3 years ago

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The first term in a sequence is 84. The pattern follows the rule add 24. What are the next four terms?

weqwewe [10]

Answer: 108, 132, 156, and 180

Step-by-step explanation:

If you need to add 24 to the first term for the next 4 terms, you would have

84 + 24 = 108

108 + 24 = 132

132 + 24 = 156

156 + 24 = 180

So your sequence would be 108, 132, 156, and 180 for the next 4 terms.

5 0

3 years ago

Customers enter the waiting line to pay for food as they leave a cafeteria on a first-come, first-served basis. The arrival rate

Elan Coil [88]

Answer:

The answer is B) 0.57.

Step-by-step explanation:

In this problem we have to apply queueing theory.

It is a single server queueing problem.

The arrival rate is $\lambda=4$ and the service rate is $\mu=7$ .

The proportion of time that the server is busy is now as the "server utilization"and can be calculated as:

$p=\frac{\lambda}{c\mu} =\frac{4}{1*7}=0.57$

where c is the number of server (in this case, one server).

3 0

3 years ago

Jane has three boxes she needs to ship. Box X weighs 33 kg more than Box Y and Box Z weighs 12 kg more than Box Y. She

andreyandreev [35.5K]

Answer:

$3w+45=557$

Step-by-step explanation:

Let the weight of box Y be w kg.

Box X weighs 33 kg more than Box Y, then the weight of box X is

$w+33$ kg.

Box Z weighs 12 kg more than Box Y, then the weight of box Z is

$w+12$ kg.

The total weight of all boxes is

$w+(w+33)+(w+12)=w+w+33+w+12=3w+45$ kg

Jane measured the total weight of all the boxes and it was found to be 557 kg, therefore,

$3w+45=557$

or

$3w=512$

3 0

3 years ago